NCERT Solutions Class 10 Maths Chapter 4

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Extramarks offers NCERT Solutions for Class 10 Mathematics Chapter 4 to help students practise the exercise questions given at the end of the Chapter in the NCERT textbook. Class 10 Mathematics chapter 4 solutions have answers written in a step-by-step manner so that students find it easier to understand the concept. The subject matter experts at Extramarks have prepared the solutions while keeping in mind the guidelines by CBSE.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations 

Access NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

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Class 10 Mathematics Chapter 4 Solutions on Extramarks have answers to all the questions given at the end of the chapter in the NCERT textbook. The answers are written in simple language and are easy to comprehend. Students can refer to these solutions to cross-check their answers, solve the questions, and for last-minute preparation.

NCERT Solutions for Class 10 Maths

Mathematics is one of the most scoring subjects and therefore your overall performance in the exams depends upon how well you have done in your Mathematics subject. Let’s look at the chapters included in Class 10 Mathematics:

  • Chapter 1 – Real Numbers 
  • Chapter 2 – Polynomials 
  • Chapter 3 – Pair of Linear Equations in Two Variables 
  • Chapter 4 – Quadratic Equations 
  • Chapter 5 – Arithmetic Progressions 
  • Chapter 6 – Triangles 
  • Chapter 7 – Coordinate Geometry 
  • Chapter 8 – Introduction to Trigonometry 
  • Chapter 9 – Some Applications of Trigonometry 
  • Chapter 10 – Circles 
  • Chapter 11 – Constructions 
  • Chapter 12 – Areas Related to Circles 
  • Chapter 13 – Surface Areas and Volumes 
  • Chapter 14 – Statistics 
  • Chapter 15 – Probability

What is the advantage of Class 10 Maths NCERT Solutions Chapter 4 by Extramarks?

There are numerous benefits of referring to NCERT Solutions for Class 10 Mathematics Chapter 4 Quadratic Equations. Let’s know about some of those advantages.

  • Detailed Answer to Problems: All the answers in NCERT Solutions for Class 10 Mathematics Chapter 4 are prepared by subject matter experts. The explanations are in-depth making it easier for students to understand the reasoning behind the answer.
  • Revision is Easy: Class 10 Mathematics Chapter 4 NCERT Solutions make it easier for students to revise their exercises anytime. Quick notes provided with the exercises help revise the important points and thus recall all the formulas, applications, and other essential points. 
  • Time-Saving for Students: With Class 10 Mathematics Chapter 4 Solutions, students need not waste time going through several study materials to find accurate answers. They will find the answers to every question in the textbook in the solutions by Extramarks. 

Related Questions 

Q1. The equation (x−705)(x−795)+800(x−750)(x−835)=0 has

  1. imaginary roots
  2. Equal roots
  3. Distinct real roots
  4. None of these

Ans. Imaginary roots

Q.1

Check whether the following are quadratic equations :(i) (x+1)2=2(x−3) (ii) x2−2x=−2(3−x)(iii)  (x−2)(x+1)=(x−1)(x+3) (iv) (x−3)(2x+1)=x(x+5)(v) (2x−1)(x−3)=(x+5)(x−1) (vi) x2+3x+1=(x−2)2(vii) (x+2)3=2x(x2−1) (viii) x3−4x2−x+1=(x−2)3

Ans.

(i)         (x+1)2=2(x−3)⇒  x2+2x+1=2x−6⇒  x2+2x+1−2x+6=0⇒x2+7=0It is of the form ax2+bx+c=0.Therefore, the given equation is a quadratic equation.(ii)         x2−2x=(−2)(3−x)⇒  x2−2x=2x−6⇒  x2−2x−2x+6=0⇒x2−4x+6=0It is of the form ax2+bx+c=0.Therefore, the given equation is a quadratic equation.(iii)         (x−2)(x+1)=(x−1)(x+3)⇒ x2−x−2=x2+2x−3⇒ x2−x−2−x2−2x+3=0⇒   −3x+1=0It is not of the form ax2+bx+c=0.Therefore, the given equation is not a quadratic equation.(iv)      (x−3)(2x+1)=x(x+5)⇒ 2x2−5x−3=x2+5x⇒ 2x2−5x−3−x2−5x=0⇒   x2−10x−3=0It is of the form ax2+bx+c=0.Therefore, the given equation is a quadratic equation.(v)      (2x−1)(x−3)=(x+5)(x−1)⇒ 2x2−7x+3=x2+4x−5⇒ 2x2−7x+3−x2−4x+5=0⇒   x2−11x+8=0It is of the form ax2+bx+c=0.Therefore, the given equation is a quadratic equation.(vi)      x2+3x+1=(x−2)2⇒x2+3x+1=x2−4x+4⇒x2+3x+1−x2+4x−4=0⇒  7x−3=0It is not of the form ax2+bx+c=0.Therefore, the given equation is not a quadratic equation.(vii)      (x+2)3=2x(x2−1)⇒x3+6x2+12x+8=2x3−2x⇒x3+6x2+12x+8−2x3+2x=0⇒  −x3+6x2+14x+8=0It is not of the form ax2+bx+c=0.Therefore, the given equation is not a quadratic equation.(viii)      x3−4x2−x+1=(x−2)3⇒ x3−4x2−x+1=x3−6x2+12x−8⇒x3−4x2−x+1−x3+6x2−12x+8=0⇒  2x2−13x+9=0It is of the form ax2+bx+c=0.Therefore, the given equation is a quadratic equation.

Q.2 Represent the following situations in the form of quadratic equations:

1. The area of a rectangular plot is 528 m­­­2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

2. The product of two consecutive positive integers is 306. We need to find the integers.

3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

4. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Ans.

(i) Let the breadth of the plot be x m. Then, as per given information, the length of the plot is (2x+1) m.            Also, given that area of the rectangular plot is 528 m2. Therefore,          (2x+1)x=528         [Area of a rectangle=Length×Breadth]    or   2x2+x−528=0.(ii) Let two consecutive positive integers are x and x+1. Then we have,        x(x+1)=306⇒    x2+x−306=0(iii) Let Rohan’s present age is x years. Then Rohan’s mother is x+26 years old. According to question,         (x+3)(x+26+3)=360⇒     (x+3)(x+29)=360⇒     x2+32x+87−360=0⇒     x2+32x−273=0(iv) Let uniform speed of the train is x km/h. Then the time  taken to travel the distance of 480 km is 480x hours. Again, when the speed is (x−8) km/h then the time taken to travel the distance of 480 km is 480x−8 hours. According to question,         480x−8−480x=3⇒     480(x−x+8)=3x(x−8)⇒     480×83=x2−8x⇒     x2−8x−1280=0

Q.3

Find the roots of the following quadratic equationsby factorisation: (i) x2−3x−10=0 (ii) 2x2+x−6=0(iii) 2x2+7x+52=0 (iv) 2x2−x+18=0(v) 100x2−20x+1=0

Ans.

(i)        x2−3x−10=0⇒  x2−5x+2x−10=0⇒  x(x−5)+2(x−5)=0⇒  (x−5)(x+2)=0⇒  x−5=0 or  x+2=0⇒  x=5 or  x=−2Therefore, roots of the equation x2−3x−10=0 are 5 and –2.(ii)        2x2+x−6=0⇒  2x2+4x−3x−6=0⇒  2x(x+2)−3(x+2)=0⇒  (2x−3)(x+2)=0⇒  2x−3=0 or  x+2=0⇒  x=32 or  x=−2Therefore, roots of the equation 2x2+x−6=0 are 32 and –2.(iii)        2x2+7x+52=0 ⇒  2x2+2x+5x+52=0⇒  2x(x+2)+5(x+2)=0⇒  (2x+5)(x+2)=0⇒  2x+5=0 or  x+2=0⇒  x=−52 or  x=−2Therefore, roots of the equation 2x2+7x+52=0are −52 and –2.(iv)        2x2−x+18=0 ⇒  16x2−8x+1=0 ⇒  16x2−4x−4x+1=0⇒  4x(4x−1)−1(4x−1)=0⇒  (4x−1)(4x−1)=0⇒  4x−1=0 or  4x−1=0⇒  x=14 or  x=14Therefore, roots of the equation 2x2−x+18=0 are 14 and 14.(v)        100x2−20x+1=0 ⇒  100x2−10x−10x+1=0⇒  10x(10x−1)−1(10x−1)=0⇒  (10x−1)(10x−1)=0⇒  10x−1=0 or  10x−1=0⇒  x=110 or  x=110Therefore, roots of the equation 100x2−20x+ 1=0 are 110 and 110.

Q.4 Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.

Ans.

(i) John and Jivanti together have 45 marbles.So, let the number of marbles John had be x.Then, number of marbles Jivanti had=45−xBoth of them lost 5 marbles each, thennumber of marbles left with John=x−5andnumber of marbles left with Jivanti=45−x−5=40−xThe product of the number of marbles they now have is 124.Therefore,       (x−5)(40−x)=124 ⇒  40x−x2−200+5x−124=0 ⇒  x2−45x+324=0⇒  x2−36x−9x+324=0⇒  x(x−36)−9(x−36)=0⇒  (x−9)(x−36)=0⇒  x−9=0 or  x−36=0⇒  x=9 or  x=36So, if John had 9 marbles then Jivanti had 36 marblesand if John had 36 marbles then Jivanti had 9 marbles.(ii)Let the number of toys produced be x.Given that cost of production of each toy is55 minus the number of toys produced in a day.∴Cost of production of each toy = ₹(55−x)Given that total cost of production of the toys in a day= ₹750Therefore,       (55−x)x=750 ⇒  −x2+55x−750=0 ⇒  x2−55x+750=0⇒  x2−25x−30x+750=0⇒  x(x−25)−30(x−25)=0⇒  (x−30)(x−25)=0⇒  x−30=0 or  x−25=0⇒  x=30 or  x=25So, number of toys is either 25 or 30.

Q.5 Find two numbers whose sum is 27 and product is 182.

Ans.

Let the numbers are x and y, then we have      x+y=27                           ...(1)and      xy=182⇒x(27−x)=182      [Putting the value of y from equation  (1)]⇒27x−x2−182=0⇒x2−27x+182=0⇒x2−13x−14x+182=0⇒x(x−13)−14(x−13)=0⇒(x−13)(x−14)=0⇒x−13=0 or x−14=0⇒x=13 or x=14Putting x=13 in equation (1), we get y=14andPutting x=14 in equation (1), we get y=13So, the two numbers are 13 and 14.

Q.6 Find two consecutive positive integers, sum of whose squares is 365.

Ans.

Let the two consecutive positive integers are x and x+1 andsum of their square, is 365.Therefore,      x2+x+12=365⇒  x2+x2+2x+1−365=0⇒2x2+2x−364=0⇒x2+x−182=0⇒x2+14x−13x−182=0⇒x(x+14)−13(x+14)=0⇒(x+14)(x−13)=0⇒x+14=0 or x−13=0⇒x=−14     or x=13Integers are positive. So, we take x=13Thus the consecutive positive integers are 13 and 14.

Q.7 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Ans.

Let the base of a right triangle is x cm. Then its altitude is(x−7) cm. It is given that hypotenuse of the triangle is 13 cm.Using Pythagoras theorem, we have         x2+x−72=132⇒ x2+x2−14x+49=169⇒ 2x2−14x+49−169=0⇒ 2x2−14x−120=0⇒ x2−7x−60=0⇒ x2−12x+5x−60=0⇒ x(x−12)+5(x−12)=0⇒ (x−12)(x+5)=0⇒ x−12=0 or x+5=0⇒ x=12 or x=−5Length can not be negative. Thus, base =x=12 cm and altitude=x−7=12−7=5 cm.

Q.8 A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Ans.

Let the number of pottery articles produced be x.Cost of production of each article is 3 more than twice the number of articles produced in a day.∴Cost of production of each article=  ₹(2x+3) Total cost of production of the rticles in a day= ₹90Therefore,       (2x+3)x=90 ⇒  2x2+3x−90=0 ⇒  2x2+15x−12x+90=0⇒  x(2x+15)−6(2x+15)=0⇒  (2x+15)(x−6)=0⇒  2x+15=0 or  x−6=0⇒  x=−152 or  x=6So, number of articles produced in a day is 6.Cost of each article =2×6+3=  ₹15

Q.9 

Find the roots of the following quadratic equations,if they exist, by the method of completing the square: (i) 2x2−7x+3=0 (ii)  2x2+x−4=0  (iii) 4x2+43x+3=0 (iv) 2x2+x+4=0

Ans.

(i) 2x2−7x+3=0⇒  x2−72x+32=0⇒  x2−2×74x+(74)2−(74)2+32=0⇒  (x−74)2−4916+32=0⇒  (x−74)2−2516=0⇒  (x−74)2−(54)2=0⇒  (x−74)2=(54)2⇒  x−74=±54⇒  x=74±54⇒  x=124=3 or   x=24=12(ii) 2x2+x−4=0⇒  x2+12x−2=0⇒  x2+2×14x+(14)2=(14)2+2⇒  (x+14)2=116+2=3316⇒  (x+14)2=(334)2⇒  x+14=±334⇒  x=−14±334⇒  x=−1+334 or   x=−1−334 (iii) 4x2+43x+3=0⇒  (2x)2+2(2x) (3) +(3)2=0⇒  (2x+3)2=0⇒  (2x+3)(2x+3)=0⇒  2x+3=0 or 2x+3=0⇒  x=−32 or x=−32 (iv) 2x2+x+4=0⇒  x2+12x+2=0⇒  x2+2×14x+(14)2−(14)2+2=0⇒  (x+14)2=116−2=−3116 Square of a number can not be negative.Therefore, there is no real root for the given equation.

Q.10

Find the roots of the following quadratic equations,by applying quadratic formula: (i) 2x2−7x+3=0 (ii)  2x2+x−4=0  (iii) 4x2+43x+3=0 (iv) 2x2+x+4=0

Ans.

(i) 2x2−7x+3=0On comparing the given equation with ax2+bx+c=0, we geta=2,  b=−7,  c=3By using quadratic formula, we get      x=−b±b2−4ac2a⇒  x=−(−7)±(−7)2−4×2×32×2⇒  x=7±49−244⇒  x=7±254=7±54⇒  x=7+54 or 7−54⇒  x=124=3 or   x=24=12(ii) 2x2+x−4=0On comparing the given equation with ax2+bx+c=0,  we geta=2,  b=1,  c=−4By using quadratic formula, we get      x=−b±b2−4ac2a⇒  x=−1±12−4×2×(−4)2×2⇒  x=−1±12+324⇒  x=−1±334⇒  x=−1+334 or   x=−1−334 (iii) 4x2+43x+3=0On comparing the given equation with ax2+bx+c=0,  we geta=4,  b=43,  c=3By using quadratic formula, we get       x=−b±b2−4ac2a⇒   x=−43±(43)2−4×4×32×4⇒   x=−43±48−488⇒  x=−43±08⇒  x=−32 or x=−32(iv) 2x2+x+4=0On comparing the given equation with ax2+bx+c=0,  we geta=2,  b=1,  c=4By using quadratic formula, we get       x=−b±b2−4ac2a⇒  x=−1±12−4×2×42×2⇒  x=−1±−314 Square of a number can not be negative.Therefore, −31 is not real and so there is no real root forthe given equation.

Q.11

Find the roots of the following equations: (i) x−1x=3, x≠0 (ii)  1x+4−1x−7=1130, x≠−4,7

Ans.

(i)        x−1x=3, x≠0⇒x2−1x=3⇒x2−1=3x⇒x2−3x−1=0On comparing this equation with ax2+bx+c=0,  we geta=1,  b=−3,  c=−1By using quadratic formula, we get       x=−b±b2−4ac2a⇒   x=−(−3)±(−3)2−4×1×(−1)2×1⇒   x=3±9+42⇒   x=3+132 or x=3−132(ii)  1x+4−1x−7=1130, x≠−4,7⇒   x−7−x−4(x+4)(x−7)=1130⇒−11x2−3x−28=1130⇒x2−3x−28=−30⇒x2−3x+2=0On comparing this equation with ax2+bx+c=0,  we geta=1,  b=−3,  c=2By using quadratic formula, we get       x=−b±b2−4ac2a⇒   x=−(−3)±(−3)2−4×1×22×1⇒   x=3±9−82⇒   x=3+12 or x=3−12⇒   x=2 or x=1

Q.12

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 13. Find his present age.

Ans.

Let the present age of Rehman is x years.​ Then 3 years ago,Rehman’s age =(x−3) years5 years from now, Rehman’s age =(x+5) yearsAccording to question,         1x−3+1x+5=13⇒     x+5+x−3(x−3)(x+5)=13⇒     2x+2x2+2x−15=13⇒     6x+6=x2+2x−15⇒     x2+2x−15−6x−6=0⇒     x2−4x−21=0On comparing this equation with ax2+bx+c=0,  we geta=1,  b=−4,  c=−21By using quadratic formula, we get       x=−b±b2−4ac2a⇒   x=−(−4)±(−4)2−4×1×(−21)2×1⇒   x=4±16+842⇒   x=4+102 or x=4−102⇒   x=7 or x=−3Age can not be a negative quantity. Therefore, Rehman’spresent age is 7 years.

Q.13 In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Ans.

Let Shefali’s marks in Mathematics and English be x and yrespectively.According to question,      x+y=30 ...(1)and      (x+2)(y−3)=210                          ...(2)⇒(x+2)(30−x−3)=210     [from (1), y=30−x]⇒(x+2)(27−x)=210  ⇒27x−x2+54−2x=210⇒25x−x2=210−54=156⇒x2−25x+156=0⇒x2−12x−13x+156=0⇒x(x−12)−13(x−12)=0⇒(x−12)(x−13)=0⇒x=12 or x=13On putting these values of x in (1), we gety=18     when x=12andy=17     when x=13Therefore, Shefali’s marks in Mathematics and English areeither 12 and 18 or 13 and 17 respectively.

Q.14

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metresmore than the shorter side, find the sides of the field.

Ans.

Let length of the shorter side is x. Then length of the longer side is (x+30)and length of the diagonal is(x+60).Using Pythagoras theorem, we have        x2+(x+30)2=(x+60)2⇒x2+x2+60x+900=x2+120x+3600⇒x2−60x−2700=0⇒x2−90x+30x−2700=0⇒x(x−90)+30(x−90)=0⇒(x−90)(x+30)=0⇒x−90=0 or x+30=0⇒x=90 or x=−30Length can not be negative.Thus, x=90 mand x+30=90+30=120 mHence, length of the shorter and longer sides of the rectangular field are 90 m and 120 m respectively.

Q.15 The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Ans.

Let smaller and larger number be x and y respectively. Then, we have         y2−x2=180and        x2=8yReplacing x2 with 8y in y2−x2=180, we get       y2−8y=180⇒y2−8y−180=0⇒y2−18y+10y−180=0⇒y(y−18)+10(y−18)=0⇒(y−18)(y+10)=0⇒y−18=0 or y+10=0⇒y=18 or y=−10Negative value of y does not satisfy the equation x2=8y.Therefore, y=18and hence         x2=8y⇒     x2=8×18=144=122⇒     x=±12Hence larger number is 18 and smaller number is either12 or −12.

Q.16 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Ans.

Let the speed of the train be x km/h.Time taken to cover 360 km is 360xhours.According to question,      (x+5)(360x−1)=360⇒(x+5)(360−x)=360x⇒360x−x2+1800−5x=360x⇒−x2+1800−5x=0⇒x2+5x−1800=0⇒x2+45x−40x−1800=0⇒x(x+45)−40(x+45)=0⇒(x+45)(x−40)=0⇒x+45=0 or x−40=0⇒x=−45       or x=40Here, speed can not be negative.Therefore, speed =x=40 km/h

Q.17

Two water taps together can fill a tank in  938 hours. The tap of larger diameter takes 10 hours less thanthe smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Ans.

Let the tap of smaller diameter fills the tank in x hours.So, in 1 hour it fills 1x part of the tank.The tap of larger diameter fills the tank in (x−10) hoursand so in 1 hour it fills 1x−10 part of the tank.The two water taps together fill the tank in 938hours.So, we have            938(1x)+938(1x−10)=1⇒       758(1x)+758(1x−10)=1⇒       758(x−10+xx(x−10))=1⇒       758(2x−10x2−10x)=1⇒       150x−750=8x2−80x⇒       8x2−230x+750=0⇒       8x2−200x−30x+750=0⇒       8x(x−25)−30(x−25)=0⇒       (8x−30)(x−25)=0⇒       8x−30=0 or   x−25=0⇒       x=308=154 or   x=25If x=154, then x−10=154−10=−254But time can not be negative. So, here we take x=25 and so we have x−10=25−10=15Therefore, the tap of smaller and larger diameters fill thetank in 25 hours and 15 hours respectively.

Q.18 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

Ans.

Let average speed of the passenger train be x km/h.Then average speed of the express train is (x + 11) km/h.It is given that the time taken by the express trainto cover 132 km is 1 hour less than the passengertrain to cover the same distance.Therefore,            132x−132x+11=1or 132(x+11−x)=x(x+11)or 1452=x2+11xor x2+11x−1452=0or x2+44x−33x−1452=0or x(x+44)−33(x+44)=0or (x−33)(x+44)=0or x−33=0 or x+44=0or x=33 or x=−44Speed can not be negative.Therefore, Average speed of the passanger train=x km/h=33 km/hAverage speed of the express train=(x+11) km/h=44 km/h

Q.19 Sum of the areas of two squares is 468 m­2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Ans.

Let sides of the two squares are x and y.According to question,4x−4y=24    ⇒   x−y=6⇒x=y+6                            ...(1)Also, we have            x2+y2=468⇒        (y+6)2+y2=468⇒        y2+12y+36+y2=468⇒        2y2+12y+36−468=0⇒        2y2+12y−432=0⇒        y2+6y−216=0⇒        y2+18y−12y−216=0⇒        y(y+18)−12(y+18)=0⇒        (y−12)(y+18)=0⇒        y−12=0 or y+18=0⇒        y=12 or y=−18Side of a square can not be negative.So, y=12Substituting this value of y in equation (1), we get           x=12+6=18Hence, the sides of the two squares are 12 m and 18 m.

Q.20

Find the nature of the roots of the following quadraticequations. If the real roots exist, find them.(i) 2x2−3x+5=0 (ii)  3x2−43x+4=0(iii) 2x2−6x+3=0

Ans.

(i) 2x2−3x+5=0 Here, a=2, b=−3, c=5 Therefore, discriminant b2−4ac=−32−4×2×5                                                                          = 9−40                                                                          =−31<0 So, the given equation has no real roots.(ii)  3x2−43x+4=0 Here, a=3, b=−43, c=4 Therefore, discriminant b2−4ac=−432−4×3×4                                                                          = 48−48                                                                          =0 So, the given equation has two equal real roots. The roots are −b2a,  −b2a,   i.e., 436,  436, i.e., 23, 23.(iii) 2x2−6x+3=0 Here, a=2, b=−6, c=3 Therefore, discriminant b2−4ac=−62−4×2×3                                                                             =36−24                                                                             =12>0 So, the given equation has two distinct real roots. The roots are given by −b+b2−4ac2a and  −b–b2−4ac2a. −b±b2−4ac2a=6±124=3±32 Hence, the roots are 3+32 and 3−32.

Q.21 Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx +3 = 0
(ii) kx(x – 2) + 6 = 0

Ans.

(i) 2x2+kx+3=0 Discriminant must be zero for two roots of the given equation to be equal. Therefore, b2−4ac=0⇒ k2−4×2×3=0⇒k2−24=0⇒k2=24⇒k=24=±26(ii)         kx(x−2)+6=0⇒kx2−2kx+6=0 Discriminant must be zero for two roots of the given equation to be equal. Therefore, b2−4ac=0⇒ 4k2−4×k×6=0⇒k2−6k=0⇒k(k−6)=0⇒k=0 or k=6But k=0 does not satisfy the given equation.Thus, the given equation has equal roots if the value of k is 6 only.

Q.22 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Ans.

Let breadth of the rectangular mango grove is x.According to question,       x×2x=800⇒x2=400⇒x2−400=0Discriminant=b2−4ac=0−4×1×(−400)=1600>0Therefore, the above equation has two distinct real roots.Now,      x2−400=0⇒x2=400⇒x=±20But breadth can not be negative.So, x=20 and 2x=40Hence, length=40 m                breadth=20 m

Q.23 Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Ans.

Let age of one friend be x years.Age of the other friend =(20−x) yearsAccording to question,       (x−4)(20−x−4)=48⇒(x−4)(16−x)=48⇒−x2+16x−64+4x−48=0⇒x2−20x+112=0Discriminant=b2−4ac=(−20)2−4×112=400−448=−48<0Therefore, no real root is possible and so the given situationis not possible.

Q.24 Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Ans.

Let length and breadth of the rectangular park be x.and y respectively.According to question, we have       2(x+y)=80⇒x+y=40⇒y=40−x                                            ...(1)Also, we have      xy=400⇒x(40−x)=400⇒40x−x2=400⇒x2−40x+400=0Discriminant=b2−4ac=(−40)2−4×400=0So, the above equation has two equal real roots.Now, x2−40x+400=0⇒(x−20)2=0⇒x=20 or x=20From equation (1), we have          y=40−x=40−20=20Hence, length and breadth of the rectangular park are20 m and 20 m respectively.

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NCERT Solutions for Class 10 Maths Related Chapters

Real Numbers

Polynomials

Pair of Linear Equations in Two Variables

Arithmetic Progressions

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Surface Areas and Volumes

Statistics

Probability

FAQs (Frequently Asked Questions)

1. What is the unit-wise mark weightage for Class 10 Mathematics?

The below chart will provide you with information about unit-wise weightage for class 10 Mathematics:

Unit I: Number Systems – 06 Marks

Unit II: Algebra – 20 Marks

Unit III: Coordinate Geometry – 06 Marks

Unit IV: Geometry – 15 Marks

Unit V: Trigonometry – 12 Marks

Unit VI: Mensuration – 10 Marks

Unit VII: Statistics & Probability – 11 Marks

2. What important topics do NCERT Class 10 Mathematics Chapter 4 Solutions contain?

Class 10 Mathematics chapter 4 NCERT solutions contain the following topics:

  • Exercise 4.1- Introduction
  • Exercise 4.2- Quadratic Equations
  • Exercise 4.3- Solution of a Quadratic Equation by Factorisation
  • Exercise 4.4- Solution of a Quadratic Equation by Completing the Square
  • Exercise 4.5- Nature of Roots

3. How and from where can I getNCERT Solutions for Class 10 Mathematics Chapter 4?

You can access NCERT Solutions for Class 10 Mathematics Chapter 4 from the official website of Extramarks.